\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 27, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/27\hfil Periodic solutions on time scales]
{Periodicity and stability in neutral nonlinear dynamic equations
with functional delay on a time scale}

\author[E. R. Kaufmann, Y. N. Raffoul\hfil EJDE-2007/27\hfilneg]
{Eric R. Kaufmann, Youssef N. Raffoul} % in alphabetical order

\address{Eric R. Kaufmann \newline
Department of Mathematics \& Statistics\\
University of Arkansas at Little Rock, Little Rock, AR 72204, USA}
\email{erkaufmann@ualr.edu}

\address{Youssef N. Raffoul \newline
Department of Mathematics\\
University of Dayton, Dayton, OH 45469-2316, USA}
\email{youssef.raffoul@notes.udayton.edu}


\thanks{Submitted July 11, 2006. Published February 12, 2007.}
\subjclass[2000]{34K13, 34C25, 34G20}
\keywords{Krasnosel'ski\u{\i}; contraction mapping; neutral;
nonlinear; Delay; \hfill\break\indent
time scales;  periodic solution; unique solution;
stability}

\begin{abstract}
 Let $\mathbb{T}$ be a periodic time scale. We use a
 fixed point theorem due to Krasnosel'ski\u{\i} to show that the
 nonlinear neutral dynamic equation with delay
 \[
 x^{\Delta}(t) = -a(t)x^{\sigma}(t)+ \left(Q(t,x(t), x(t-g(t))))\right)^{\Delta}
 + G\big(t,x(t), x(t-g(t))\big), t \in \mathbb{T},
 \]
 has a periodic solution.  Under a slightly
 more stringent inequality we show that the periodic solution is
 unique using the contraction mapping principle. Also, by the aid
 of the contraction mapping principle we study the asymptotic stability of
 the zero solution provided that $Q(t,0,0)= G(t,0,0) = 0$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}

\section{Introduction}

We assume the reader is familiar with the notation and basic results
for dynamic equations on time scales. For a review of this topic we
direct the reader to the monographs \cite{mbap, mbap2}. We begin
with a few definitions.

\begin{definition}\label{def1.1} \rm
We say that a time scale $\mathbb{T}$ is {\em periodic} if there
exist a $p > 0$ such that if $t \in \mathbb{T}$ then $t \pm p \in
\mathbb{T}$. For $\mathbb{T} \neq \mathbb{R}$, the smallest positive
$p$ is called the \emph{period} of the time scale.
\end{definition}

The above definition is due to Atici \emph{et. al} \cite{agk} and
Kaufmann and Raffoul \cite{er2}.

\begin{example} \label{exa1.2} \rm
The following time scales are periodic.
\begin{enumerate}
    \item $\mathbb{T} =  \bigcup_{i =
    -\infty}^{\infty} [2(i-1)h, 2ih], \, h > 0$ has period $p =
    2h$.

    \item $\mathbb{T} = h\mathbb{Z}$ has period $p = h$.

    \item $\mathbb{T} = \mathbb{R}$.

    \item $\mathbb{T} = \{ t = k - q^m : k \in \mathbb{Z}, m \in
    \mathbb{N}_0\}$ where,, $0 < q < 1$ has period $p = 1$.

\end{enumerate}
\end{example}

\noindent \textbf{Remark:} All periodic time scales are unbounded
above and below.

\begin{definition} \label{def1.3} \rm
Let $\mathbb{T} \neq \mathbb{R}$ be a periodic time scale with
period $p$. We say that the function $f\colon \mathbb{T} \to
\mathbb{R}$ is periodic with period $T$ if there exists a natural
number $n$ such that $T = np$, $f(t \pm T) = f(t)$ for all $t \in
\mathbb{T}$ and $T$ is the smallest number such that $f(t \pm T) =
f(t)$.

If $\mathbb{T} = \mathbb{R}$, we say that $f$ is periodic with
period $T > 0$ if $\,T$ is the smallest positive number such that
$f(t \pm T) = f(t)$ for all $t \in \mathbb{T}$.
\end{definition}

\noindent \textbf{Remark:} If $\mathbb{T}$ is a periodic time scale
with period $p$, then $\sigma(t \pm np) = \sigma(t) \pm np$.
Consequently, the graininess function $\mu$ satisfies $\mu(t \pm np)
= \sigma(t \pm np) - (t \pm np) = \sigma(t) - t = \mu(t)$ and so, is
a periodic function with period $p$.

Let $\mathbb{T}$ be a periodic time scale such that $0 \in
\mathbb{T}$. We will show the existence of periodic solutions for
the nonlinear neutral dynamic equation with delay
\begin{equation}\label{eq1}
 x^{\Delta}(t) = -a(t)x^{\sigma}(t)+ \left(Q(t,x(t), x(t-g(t))))\right)^{\Delta}+
 G\big(t,x(t), x(t-g(t))\big), t \in \mathbb{T},
\end{equation}
 In order for the function $x(t - g(t))$ to
be well-defined over $\mathbb{T}$, we assume that $g : \mathbb{T}
\to \mathbb{R}$ and that $id - g : \mathbb{T} \to \mathbb{T}$ is
strictly increasing.

In the case $\mathbb{T} = \mathbb{R}$, the second author in
\cite{r} used Krasnosel'ski\u{\i}'s fixed point theorem to show
the existence of a periodic solution of (\ref{eq1}) when the delay
is some positive continuous and periodic function $g(t)$. Also,
the existence of a unique periodic solution of (\ref{eq1}) was
obtained by the aid of the contraction mapping principle. For
$\mathbb{T} = \mathbb{R}$ and assuming $x= 0$ is a solution of
(\ref{eq1}), in \cite{r1}, Raffoul used the notion of fixed point
theory and obtained conditions that guaranteed the asymptotic
stability of a zero solution.

In Section 2, we present some preliminary material that we will need
through the remainder of the paper. We will sate some facts about
the exponential function on a time scale as well as the
Krasnosel'ski\u{\i} fixed point theorem. We present our main results
on periodicity in Section 3 and provide an example. In Section 4 we
state and prove a theorem concerning the stability of the zero
solution of \eqref{eq1} for a general time scale.

\section{Preliminaries}

We begin this section by considering some advanced topics in the
theory of dynamic equations on time scales. Most of the following
definitions, lemmas and theorems can be found in \cite{mbap,
mbap2}. Our first two theorems concern the composition of two
functions. The first theorem is the chain rule on time scales
\cite[Theorem 1.93]{mbap}.

\begin{theorem}[Chain Rule]\label{thm2.1}
Assume $\nu: \mathbb{T} \to \mathbb{R}$ is strictly increasing and
$\tilde{\mathbb{T}} := \nu(\mathbb{T})$ is a time scale. Let $w :
\tilde{\mathbb{T}} \to \mathbb{R}$. If $\nu^\Delta (t)$ and
$w^{\tilde{\Delta}}(\nu(t))$ exist for $t \in \mathbb{T}^\kappa$,
then
\[
    (w \circ \nu)^\Delta = (w^{\tilde{\Delta}} \circ \nu)\nu^\Delta.
\]
\end{theorem}

In the sequel we will need to differentiate and integrate functions
of the form $f(t-g(t)) = f(\nu(t))$ where, $\nu(t) := t - g(t)$. Our
second theorem is the substitution rule \cite[Theorem 1.98]{mbap}.

\begin{theorem}[Substitution]\label{thm2.2}
Assume $\nu : \mathbb{T} \to \mathbb{R}$ is strictly increasing
and $\tilde{\mathbb{T}} := \nu(\mathbb{T})$ is a time scale. If $f
: \mathbb{T} \to \mathbb{R}$ is an rd-continuous function and
$\nu$ is differentiable with rd-continuous derivative, then for
$a, b \in \mathbb{T}$,
\[
    \int_a^b \! f(t) \nu^\Delta(t) \, \Delta t =
    \int_{\nu(a)}^{\nu(b)} (f \circ \nu^{-1}) (s) \,
    \tilde{\Delta}s.
\]
\end{theorem}

A function $p : \mathbb{T} \to \mathbb{R}$ is said to be {\em
regressive} provided $1 + \mu(t) p(t) \neq 0$ for all $t \in
\mathbb{T}^\kappa$. The set of all regressive rd-continuous
functions $f : \mathbb{T} \to \mathbb{R}$ is denoted by
$\mathcal{R}$ while the set $\mathcal{R}^+$ is given by
$\mathcal{R}^+ = \{f \in \mathcal{R} : 1 + \mu(t) f(t) > 0 \mbox{
for all } t \in \mathbb{T} \}$.

Let $p \in \mathcal{R}$ and $\mu(t) \ne 0$ for all $t \in \mathbb{T}$. The \emph{exponential
function} on $\mathbb{T}$ is defined by
\[\label{eq2}
    e_p(t,s) = \exp \Big( \int_s^t \! \frac{1}{\mu(z)}\mbox{Log}( 1 + \mu(z)p(z)) \,
    \Delta z \Big),
\]
It is well known that if $p \in \mathcal{R}^+$, then $e_p(t,s)
> 0$ for all $t \in \mathbb{T}$. Also, the exponential function
$y(t) = e_p(t,s)$ is the solution to the initial value problem
$y^\Delta = p(t) y, \, y(s) = 1$. Other properties of the
exponential function are given in the following lemma,
\cite[Theorem 2.36]{mbap}.

\begin{lemma}\label{lemma2.3}
Let $p, q \in \mathcal{R}$. Then
\begin{itemize}
    \item[(i)] $e_0(t,s) \equiv 1$ and $e_p(t, t) \equiv 1$;
    \item[(ii)] $e_p(\sigma(t), s) = (1 + \mu(t) p(t)) e_p(t, s)$;
    \item[(iii)] $\frac{1}{e_p(t,s)} = e_{\ominus p}(t, s)$ where,
    $\ominus p(t) = - \frac{p(t)}{1 + \mu(t) p(t)}$;
    \item[(iv)] $e_p(t, s) = \frac{1}{e_p(s,t)} = e_{\ominus p}(s, t)$;
    \item[(v)] $e_p(t,s) e_p(s, r) = e_p(t,r)$;
    \item[(vi)] $\big( \frac{1}{e_p( \cdot, s)} \big) ^\Delta = -
    \frac{p(t)}{e_p^\sigma(\cdot, s)}$.
\end{itemize}
\end{lemma}

Lastly in this section, we state Krasnosel'ski\u{\i}'s fixed point
theorem which enables us to prove the existence of a periodic
solution. For its proof we refer the reader to \cite{s1}.

\begin{theorem}[Krasnosel'ski\u{\i}] \label{thm2.4}
Let $\mathbb{M}$ be a closed convex nonempty subset of a Banach
space $\big ( \mathbb{B}, \| \cdot\| \big )$. Suppose that $A$ and
$B$ map $\mathbb{M}$ into $\mathbb{B}$ such that
\begin{itemize}
\item[(i)] $x,y \in \mathbb{M}$, implies $Ax + By \in \mathbb{M}$,
\item[(ii)] $A$ is compact and continuous, \item[(iii)] $B$ is a
contraction mapping.
\end{itemize}
Then there exists $z \in \mathbb{M}$ with $z=Az+Bz$.
\end{theorem}

\section{Existence Of Periodic Solutions}

We will state and prove our main result in this section as well as
provide an example. Let $T > 0, \;T \in \mathbb{T}$ be fixed and if
$\mathbb{T} \neq \mathbb{R}$, $T= np$ for some $n \in \mathbb{N}$.
By the notation $[a, b]$ we mean
\[
    [a, b] = \{t \in \mathbb{T} : a \leq t \leq b \}
\]
unless otherwise specified. The intervals $[a,b), (a, b]$, and $(a,
b)$ are defined similarly. Define $P_T = \lbrace \varphi \in
C(\mathbb{T},R): \varphi (t+T)=\varphi (t) \rbrace$ where,
$C(\mathbb{T},R)$ is the space of all real valued continuous
functions on $\mathbb{T}$. Then $P_T$ is a Banach space when it is
endowed with the supremum norm
$$
\| x \|=\sup_{t \in [0,T]}|x(t)|.
$$

We will need the following lemma whose proof can be found in
\cite{er2}.

\begin{lemma} \label{lem3.1}
Let $x \in P_T$. Then $\|x^\sigma\|$ exists and $\|x^\sigma\| =
\|x\|$.
\end{lemma}

In this paper we assume that $a \in \mathcal{R}^+$ is continuous,
$a(t)> 0$ for all $t \in \mathbb{T}$ and
\begin{equation}\label{e3.01}
    a(t+T)=a(t),\quad  \big(id - g \big )(t+T) = \big(id - g \big
    )(t),
\end{equation}
where, \emph{id} is the identity function on $\mathbb{T}$. We also
require that $Q(t,x)$ and $G(t,x,y)$ are continuous and periodic in
$t$ and Lipschitz continuous in $x$ and $y$. That is,
\begin{equation}\label{e3.02}
   Q(t+T,x) = Q (t,x), \; G(t + T, x, y) = G(t, x, y),
\end{equation}
and there are positive constants $E_1, E_2,
 E_3$ such that  \begin{equation}\label{e3.03} |Q(t,x) -
Q(t,y)| \le E_1 \|x-y\|,\; \mbox{for}\; x, y \in \mathbb{R},
 \end{equation} and
 \begin{equation}\label{e3.04}
|G(t,x, y)- G(t,z, w)| \le E_2 \|x-z\| + E_3 \|y-w\|,\; \mbox{for}\; x, y, z, w \in \mathbb{R}.
\end{equation}

\begin{lemma} \label{lemma3.1}
Suppose \eqref{e3.01}, \eqref{e3.02} hold. If  $x\in P_T$, then
$x$ is a solution of equation \eqref{eq1} if, and only if,
\begin{equation}\label{e3.05}
\begin{aligned}
    x(t)& = Q\big(t, x(t-g(t))\big) +  \big ( 1 - e_{\ominus a}(t, t-T) \big )^{-1}\\
    & \quad \times \int_{t-T}^t \! \Big[-a(s)Q^\sigma\big(s, x(s-g(s))\big) + G\big(s,
    x(s),x(s-g(s))\big) \Big]e_{\ominus a}(t,s) \, \Delta s.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Let $x \in P_{T}$ be a solution of \eqref{eq1}.  First we write
\eqref{eq1} as
\begin{align*}
    \{x(t)-Q\big(t, x(t-g(t))\big)\}^\Delta
    &= -a(t)\{x^\sigma(t)-Q^\sigma\big(t, x(t-g(t))\big)\}\\
    &\quad -a(t)Q^\sigma\big(t,x(t-g(t))\big) + G\big(t, x(t),x(t-g(t))\big).
\end{align*}
Multiply both sides  by $e_a(t,0)$ and then integrate from $t-T$ to
$t$ to obtain
\begin{align*}
    &\int^t_{t-T} \! \big[ e_{a}(s,0)\{x(s)-Q\big(s,
      x(s-g(s))\big)\}\big ]^\Delta \, \Delta s\\
    & = \int^t_{t-T}\Big[-a(s)Q^\sigma\big(s, x(s-g(s))\big)
    + G\big(s, x(s),x(s-g(s))\big) \Big]e_{a}(s,0) \, \Delta s.
\end{align*}
Consequently, we have
\begin{align*}
    &e_{a}(t,0)\Big(x(t)-a(t)Q\big(t, x(t-g(t))\big)\Big)\\
    &  - e_{a}(t-T,0)\Big(x(t-T)
    -a(t-T)Q\big(t-T, x(t-T-g(t-T))\big)\Big)\\
    &  = \int_{t-T}^t \! \big[-a(s)Q^\sigma\big(s, x(s-g(s))\big)
    + G\big(s,x(s),x(s-g(s))\big) \Big]e_{a}(s,0) \, \Delta s.
\end{align*}
After making use of \eqref{e3.01}, \eqref{e3.02} and $x \in P_T$,
we divide both sides of the above equation by $e_a(t,0)$ to obtain
\begin{equation*}
\begin{aligned}
    x(t) &= Q\big(t, x(t-g(t))\big) +  \big ( 1 - e_{\ominus a}(t, t-T) \big )^{-1}\\
    & \quad  \times \int_{t-T}^t \! \Big[-a(s)Q^\sigma\big(s, x(s-g(s))\big)
    + G\big(s,x(s),x(s-g(s))\big) \Big]e_{\ominus a}(t,s) \, \Delta s
\end{aligned}
\end{equation*}
where, we have used Lemma \ref{lemma2.3} to simplify the
exponentials. Since each step is reversible, the converse follows.
This completes the proof.
\end{proof}

Define the mapping $H : P_{T}\to P_{T}$ by
\begin{equation}\label{e3.06}
\begin{aligned}
   & (H\varphi)(t)\\
    & =Q\big(t, \varphi(t-g(t))\big) +
    \big ( 1 - e_{\ominus a}(t, t-T) \big )^{-1}\\
   &\quad \times \int_{t-T}^t  \Big [ -a(s)Q^\sigma\big(s,
    \varphi(s-g(s))\big)
    + G\big(s, \varphi(s),\varphi(s-g(s))\big) \Big]
    e_{\ominus a}(t,s) \, \Delta s.
\end{aligned}
\end{equation}
To apply Theorem \ref{thm2.4} we need to construct two mappings;
one map is a contraction and the other map is compact. We express
equation \eqref{e3.06} as
$$
(H\varphi)(t)=(B\varphi)(t)+(A\varphi)(t)
$$
where, $A,B$ are given by
\begin{equation}\label{e3.07}
(B\varphi)(t)=Q\big(t, \varphi(t-g(t))\big)
\end{equation}
and
\begin{equation}\label{e3.08}
\begin{aligned}
&(A\varphi)(t) \\
&=  \big ( 1 - e_{\ominus a}(t, t-T) \big )^{-1}\\
& \quad \times \int_{t-T}^t \! \big[-a(s)Q^\sigma\big(s, \varphi(s-g(s))\big)
    + G\big(s,\varphi(s),\varphi(s-g(s))\big) \Big]e_{\ominus a}(t,s)
    \, \Delta s.
\end{aligned}
\end{equation}

\begin{lemma} \label{lemma3.2}
Suppose \eqref{e3.01}--\eqref{e3.04} hold. Then $A:P_{T}\to
P_{T}$, as defined by \eqref{e3.08}, is compact.
\end{lemma}

\begin{proof}
We first show that $A:P_T \to P_T$. Evaluate \eqref{e3.08} at $t +
T$.
\begin{equation}\label{e3.09}
\begin{aligned}
    (A\varphi)(t+T) &= \big ( 1 - e_{\ominus a}(t+T, t) \big
    )^{-1} \times \int_{t}^{t+T} \Big [-a(s)Q^\sigma\big(s,
    \varphi(s-g(s))\big)\\
    & \quad + G\big(s,\varphi(s),\varphi(s-g(s))\big)
    \Big] e_{\ominus a}(t+T,s) \, \Delta s.
\end{aligned}
\end{equation}
Use Theorem \ref{thm2.2} with $u = s-T$ and conditions
\eqref{e3.01} -- \eqref{e3.02} to get
\begin{align*}
    (A\varphi)(t+T)&= \big ( 1 - e_{\ominus a}(t+T, t) \big
    )^{-1}\\
    &\quad \times \int_{t-T}^{t} \Big [-a(u+T) Q^\sigma
    \big(u-T, \varphi(u-T-g(u-T))\big)\nonumber\\
    & \quad + G\big(s,\varphi(u-T),\varphi(u-T-g(u-T))\big) \Big]
    e_{\ominus a}(t+T,u+T) \, \Delta u.
\end{align*}
 From \eqref{eq2} and Theorem \ref{thm2.2}, we have $e_{\ominus
a}(t+T,u+T) = e_{\ominus a}(t, u)$ and $e_{\ominus a}(t+T,t) =
e_{\ominus a}(t, t-T)$. Thus \eqref{e3.09} becomes
\begin{align*}
    (A\varphi)(t+T)
    & = \big ( 1 - e_{\ominus a}(t, t-T) \big
    )^{-1} \times \int_{t-T}^{t} \Big [ -a(u)Q^\sigma\big(u,
    \varphi(u-g(u))\big)\\
    & \quad + G\big(u,\varphi(u),\varphi(u-g(u))\big) \Big]
    e_{\ominus a}(t+T,u) \, \Delta u\\
    & = (A\varphi)(t).
\end{align*}
That is, $A: P_T \to P_T$.

To see that $A$ is continuous, we let $\varphi,\psi\in P_{T}$ with
$\|\varphi\|\leq C$ and $\|\psi\|\leq C$ and define
\begin{equation}\label{e3.10}
  \eta := \max_{t \in [0,T]}
  \big |\big(1-e_{\ominus a}(t, t-T)\big)^{-1}\big |,\quad
  \rho := \max_{t \in [0,T]}|a(t)|,\quad
  \gamma := \max_{u \in [t-T,t]}  e_{\ominus a}(t, u).
\end{equation}
Given $\varepsilon > 0$, take $\delta =\varepsilon/M$ with $M =
\eta\;\gamma\;T(\rho\;E_1 + E_2 + E_3)$ where, $E_1$, $E_2$ and
$E_3$ are given by \eqref{e3.03} and \eqref{e3.04} such that
$\|\varphi-\psi\|<\delta$. Using \eqref{e3.08} we get
\begin{align*}
    \big \| A\varphi - A\psi \big \|
    &\leq    \eta \gamma\int^T_{0}\Big[\rho\;E_1\|\varphi-\psi\|+(E_2 +
    E_3)\|\varphi-\psi\| \Big]\Delta u\\
    &\leq M\|\varphi-\psi\|<\varepsilon.
\end{align*}
This proves that $A$ is continuous.

We need to show that $A$ is compact. Consider the sequence of
periodic functions $\big \{ \varphi_{n} \big \} \subset P_{T}$ and
assume that the sequence is uniformly bounded. Let $R > 0$ be such
that $\|\varphi_{n}\|\leq R$, for all $n \in \mathbb{N}$. In view of
\eqref{e3.03}  and \eqref{e3.04}  we arrive at
\begin{align*}
|Q(t,x)| &=|Q(t,x)- Q(t,0)+ Q(t,0)|\\
&\leq |Q(t,x)- Q(t,0)|+ |Q(t,0)|\\
&\leq E_1\|x\| + \alpha.
\end{align*}
Similarly,
\begin{align*}
|G(t,x,y)| &= |G(t,x,y)- G(t,0,0)+ G(t,0,0)|\\
&\leq |G(t,x,y)- G(t,0,0)|+ |G(t,0,0)|\\
&\leq E_2\|x\| + E_3\|y\| + \beta,
\end{align*}
where, $\alpha =  \sup_{t\in[0,T]}|Q(t,0)|$ and
$\beta =\sup_{t\in[0,T]}|G(t,0,0)|$.
\begin{align*}
    |(A\varphi_n)(t)|
& = \Big | \big ( 1 - e_{\ominus a}(t,
    t-T) \big )^{-1} \times \int_{t-T}^t \Big [-a(s)Q^\sigma
    \big(s, \varphi(s-g(s))\big)\\
&\quad + G\big(s, \varphi(s),\varphi(s-g(s))\big) \Big] e_{\ominus a}(t,s) \,
    \Delta s \Big|\\
& \leq \eta \gamma \int_{t-T}^t \big | -a(s) \big | \,
    \big | Q^\sigma\big(s, \varphi(s-g(s))\big)\big | +
    \big | G\big(s,\varphi(s),\varphi(s-g(s))\big) \big | \, \Delta s\\
& \leq \eta \gamma T \big[ \rho (E_1 R + \alpha ) +
    (E_2 + E_3 ) R + \beta \big] \equiv D.
\end{align*}
Thus the sequence $\{A \varphi_n \}$ is uniformly bounded. Now, it
can be easily checked that
\[
    (A\varphi_{n})^\Delta(t) =-a(t)(A\varphi_{n})^\sigma(t)
    - a(t)Q^\sigma\big(t, \varphi(t-g(t))\big) +
    G\big(t,\varphi(t),\varphi(t-g(t))\big).
\]
Consequently,
\[
    |(A\varphi_n)^\Delta (t)| \leq D \rho + \rho(E_1 R + \alpha )
    + (E_2 + E_3 ) R + \beta
\]
for all $n$. That is $\|(A\varphi_{n})^\Delta\|\leq F$, for some
positive constant $F$. Thus the sequence $\{A\varphi_{n}\}$ is
uniformly bounded and equi-continuous. The Arzel$\grave{a}$-Ascoli
theorem implies that there is a subsequence $\{A\varphi_{{n}_{k}}\}$ which  converges uniformly to
a continuous $T$-periodic function $\varphi^*$. Thus A is compact.
\end{proof}

\begin{lemma} \label{lemma3.4}
Let $B$ be defined by \eqref{e3.07} and
\begin{equation}\label{e3.11}
    E_1 \leq \zeta <1.
\end{equation}
Then $B: P_T \to P_T$ is a contraction.
\end{lemma}

\begin{proof}
Trivially, $B:P_T \to P_T$. For $\varphi, \psi \in P_{T}$,  we
have
\begin{align*}
    \| B\varphi - B\psi \|
    & = \sup_{t\in[0,T]} |B\varphi(t) -    B\psi(t)|\\
    &\leq  E_1 \sup_{t\in[0,T]} |\varphi(t-g(t))-\psi(t-g(t))|\\
    &\leq\zeta\|\varphi-\psi\|.
\end{align*}
Hence $B$ defines a contraction mapping with contraction constant
$\zeta$.
\end{proof}

\begin{theorem}\label{thm3.5}
Suppose the hypothesis of Lemma \ref{lemma3.1} hold. Suppose
\eqref{e3.01}--\eqref{e3.04} hold. Let $\alpha =
\sup_{t\in[0,T]}|Q(t,0)|$ and $\beta =\sup_{t\in[0,T]}|G(t,0,0)|$.
Let $J$ be a positive constant satisfying the inequality
\begin{equation}\label{e3.12}
\alpha+ E_1 J + \eta T \gamma \Big[\rho\;(E_1 J +\alpha) +(E_2 +
E_3)J+\beta\Big] \leq J.
\end{equation}
Let $\mathbb M = \{ \varphi \in P_{T}: \|\varphi \| \leq J\}$.
Then  (\ref{eq1}) has a
solution in $M$.
\end{theorem}

\begin{proof} Define $\mathbb M = \{ \varphi \in P_{T}:
\|\varphi \| \leq  J\}$. By Lemma \ref{lemma3.2}, $A$ is
continuous and $A \mathbb{M}$ is contained in a compact set. Also,
from Lemma \ref{lemma3.4}, the mapping $B$ is a contraction and it
is clear that $B:P_{T}\to P_{T}$ . Next, we show that if
$\varphi, \psi \in \mathbb M $, we have $\|A\varphi +B\psi\|\leq
J$. Let $\varphi, \psi \in \mathbb M $ with $\|\varphi\|,
\|\psi\|\leq J$. Then
\begin{align*}
    &\|A\varphi+B\psi\|\\
    &\leq E_1 \|\psi\| + \alpha +  \eta \, \gamma
    \int^T_0 \!\big[|a(u)|(\alpha +E_1\|\varphi\|)+E_2
    \|\varphi\| + E_3 \|\varphi\| +\beta \big]\Delta u\\
    &\leq \alpha+ E_1 J + \eta T \gamma\;\Big[\rho
    (E_1 J +\alpha) +(E_2 + E_3)J+\beta\Big]  \leq J .
 \end{align*}
We now see that all the conditions of Krasnosel'ski\u{\i}'s theorem
are satisfied. Thus there exists a fixed point $z$ in $\mathbb M$
such that $z=Az+Bz$. By Lemma \ref{lemma3.1}, this fixed point is a solution of
(\ref{eq1}). Hence  (\ref{eq1}) has a $T$-periodic solution.
\end{proof}

\begin{theorem}\label{thm3.6}
Suppose \eqref{e3.01}--\eqref{e3.04} hold. If
$$
E_1 + \eta\;\gamma\;T(\rho\;E_1 + E_2 + E_3)< 1,
$$
then  (\ref{eq1}) has a unique $T$-periodic solution.
\end{theorem}

\begin{proof}
Let the mapping $H$ be given by \eqref{e3.06}.
For $\varphi,\psi\in P_{T}$, we have
\[
    \|H\varphi-H\psi\| \leq \Big(E_1 +
    \eta\;\gamma\;T(\rho\;E_1 + E_2 + E_3)\Big)\|\varphi-\psi\|.
\]
This completes the proof by invoking the contraction mapping
principle.
\end{proof}

It is worth noting that Theorems \ref{thm3.5} and \ref{thm3.6} are
not applicable to functions such as
\[
    G\big(t,\varphi(t),\varphi(t-g(t))\big) = f_1(t)\varphi^2(t)+
    f_2(t)\varphi^2(t-g(t)),
\]
where, $f_1(t), f_2(t)$ and $g(t) > 0$ are continuous and periodic
on some applicable time scale. To accommodate such functions, we
state the following corollary, in which the functions $G$ and $Q$
are required to satisfy local Lipschitz conditions. We note that
conditions (3.2) and (3.4) require that the functions $G$ and $Q$ be
globally Lipschitz. \vskip .2cm

\begin{corollary}\label{coro3.7}
Suppose \eqref{e3.01}--\eqref{e3.04} hold and let $\alpha$ and $
\beta $ be the constants defined in Theorem \ref{thm3.5}. Let $J$ be
a positive constant and define $\mathbb M = \{ \varphi \in P_{T}:
\|\varphi \| \leq J\}$. Suppose there are positive constants
$E_1^{*}, E_2^{*}$ and $E_3^{*}$ so that for $x, y, z$ and $w  \in
\mathbb{M}$ we have
 \begin{gather*}
|Q(t,x) -Q(t,y)| \le E_1^{*} \|x-y\|,
\\
|G(t,x, y)- G(t,z, w)| \le E_2^{*} \|x-z\| + E_3^{*} \|y-w\|,
\end{gather*}
and
\begin{equation}\label{e3.13}
\alpha+ E_1^{*} J + \eta T \gamma \Big[\rho\;(E_1^{*} J +\alpha)
+(E_2^{*} + E_3^{*})J+\beta\Big] \leq J.
\end{equation}
Then  (\ref{eq1}) has a solution in $\mathbb{M}$. If in
addition,
$$
E_1^{*} + \eta\;\gamma\;T(\rho\;E_1^{*} + E_2^{*} + E_3^{*})< 1,
$$
then the solution in $\mathbb{M}$ is unique.
\end{corollary}

\begin{proof}
Let $\mathbb M = \{ \varphi \in P_{T}:\|\varphi \| \leq  J\}$.
Let the mapping $H$ be given by \eqref{e3.06}. Then the results
follow immediately from Theorem \ref{thm3.5} and Theorem \ref{thm3.6}.
\end{proof}

Now we give an example.

\begin{example} \label{exa3.8} \rm
Let $\mathbb{T}$ be a periodic time scale.  For small positive
$\varepsilon_1$ and $\varepsilon_2$, we consider the perturbed van
der Pol equation
\begin{equation}\label{e3.14}
    x^{\Delta}= -2 x^\sigma(t) + \Big(\varepsilon_1
    b(t)x^2(t-g(t)) \Big)^\Delta + \varepsilon_2 \Big(c(t)
    + x^2(t)\Big),
\end{equation}
where
\begin{equation}\label{e3.15}
    b(t+T)=b(t),\quad  \big ( id - g \big )(t+T) = \big ( id - g
    \big )(t), \quad \mbox{and} \quad c(t+T)=c(t).
\end{equation}
Also, we assume that the functions $b, c$ and $g$ are continuous
with $id - g: \mathbb{T} \to \mathbb{T}$ strictly increasing. So
we have
$$
a(t) =2, \quad Q(t, x(t-g(t)))= \varepsilon_1\;b(t)x^2(t-g(t))
$$
and
$$
G(t,x(t), x(t-g(t)))=\varepsilon_2 \Big(c(t)+x^2(t)\Big).
$$
Define $\mathbb M = \{ \varphi \in P_{T}: \|\varphi \| \leq J\}$,
where $J$ is a positive constant. Then for
 $ \varphi, \psi \in \mathbb{M}$ we have
$$
\eta = \Big(1 -  e_{\ominus 2}(t, t-T)\Big)^{-1},\quad
 \rho =3, \quad  \gamma \leq 1.
$$
Let
\[
    \iota = \sup_{t\in[0,T]}|b(t)|, \quad
 \kappa =  \sup_{t\in[0,T]}|c(t)|.
\]
Then
$$
\alpha = 0,\quad
\beta =\varepsilon_2 \kappa,\quad
E_1^{*} = \varepsilon_1 \iota J,\quad
E_3^{*}= 0, \quad
E_2^{*} =\varepsilon_2J.
$$
Thus, inequality \eqref{e3.13} becomes
\begin{equation*}
    \varepsilon_1 \iota J^2 +\eta T\Big[3\varepsilon_1
    J^2+\varepsilon_2J^2 +\varepsilon_2 \kappa\Big] \leq J
\end{equation*}
which is satisfied for small $\varepsilon_1$ and
$\varepsilon_2$. Hence,  \eqref{e3.14} has a $T$-periodic
solution, by Corollary \ref{coro3.7}. Moreover, if
\[
    \varepsilon_1 \iota J  + \eta \gamma T
    \Big[3\varepsilon_1 \iota J + \varepsilon_2 J \Big] < 1
\]
is satisfied for small $\varepsilon_1$ and $\varepsilon_2$, then
\eqref{e3.14} has a unique $T$-periodic solution.
\end{example}

\section{Stability}

Lyapunov functions and functionals have been successfully used to
obtain boundedness, stability and the existence of periodic
solutions of differential equations, differential equations with
functional delays and functional differential equations. When
studying differential equations with functional delays using
Lyapunov functionals, many difficulties arise if the delay is
unbounded or if the differential equation in question has
unbounded terms. In \cite{r1}, the second author studied, via
fixed point theory, the asymptotic stability of the zero solution
of the scalar neutral differential equation
\begin{equation}\label{e4.1}
 x^\prime(t) = -a(t)x(t)+ c(t)x^\prime(t-g(t))+
q\big(t,x(t), x(t-g(t)\big),
\end{equation}
where, $a(t), b(t), g(t)$ and $q$ are continuous in their respective
arguments. It is clear that  \eqref{eq1} is more general than
\eqref{e4.1}.

This section is mainly concerned with the asymptotic stability of
the zero solution of \eqref{eq1}. We assume that the functions $Q$
and $G$ are continuous, as before. Also, we assume that $g(t)$ is
continuous and  $g(t)\geq g^{*} > 0$, for all $t \in \mathbb{T}$
such that $t \geq t_0$ for some $t_0 \in \mathbb{T}$ and that
$Q(t,0) = G(t,0,0) = 0$ and $Q$ and $G$ obey the Lipschitz
conditions \eqref{e3.03} and \eqref{e3.04}. The techniques used in
this section are adapted from the paper \cite{r1}.

As before, we assume a time scale, $\mathbb{T}$, that is unbounded
above and below and that $ 0 \in \mathbb{T}$. Also, we assume that
$g : \mathbb{T} \to \mathbb{R}$ and that $id - g : \mathbb{T} \to
\mathbb{T}$ is strictly increasing.

To arrive at the correct mapping, we rewrite \eqref{eq1} as in the
proof of Lemma \ref{lemma3.1}, multiply both sides by $e_a(t,0)$
and then integrate from $0$ to $t$ to obtain

\begin{equation}\label{e4.2}
\begin{aligned}
    x(t)&= Q(t, x(t-g(t)))+ \Big[x(0) - Q\big(0,
    x(-g(0))\big)\Big]e_{\ominus a}(0,t)\\
    &\quad + \int^t_{0}\Big[-a(u)Q^\sigma\big(u,x(u-g(u))\big)
    + G\big(u, x(u),x(u-g(u))\big) \Big] e_{\ominus a}(t,u) \, \Delta u.
\end{aligned}
\end{equation}
Thus, we see that $x$ is a solution of \eqref{eq1} if and only if it
satisfies \eqref{e4.2}.

Let $\psi :(-\infty , 0]_{\mathbb{T}} \to \mathbb{R}$
be a given $\Delta$-differentiable bounded initial function. We
say $x(t) := x(t,0,\psi)$ is a solution of \eqref{eq1} if $x(t) =
\psi(t) $ for $t \leq 0 $ and satisfies \eqref{eq1} for $t \geq
0$. We say the zero solution of \eqref{eq1} is stable at $t_0$ if
for each $\varepsilon > 0$, there is a $\delta = \delta(
\varepsilon)
> 0$ such that $\big[\psi:(-\infty, t_{0}]_\mathbb{T} \to
\mathbb{R}$ with $\|\psi \| < \delta \big ]$ implies
$|x(t,t_0,\psi)| < \varepsilon$.

Let $C_{rd} = C_{rd}(\mathbb{T}, \mathbb{R})$ be the space of all
rd-continuous functions from $\mathbb{T} \to \mathbb{R}$
and define the set $S$ by
\begin{equation*}
    S = \left \{ \varphi \in C_{rd} :
    \, \varphi(t) = \psi(t) \mbox{ if } t \leq 0, \, \varphi(t)
    \to 0 \mbox{ as }\; t \to \infty, \mbox{ and
    } \varphi \mbox{ is bounded} \right \}.
\end{equation*}
Then, $\big(S, \|\cdot\| \big)$ is a complete metric space where,
$\|\cdot\|$ is the supremum norm. For the next theorem we impose the
following conditions.
\begin{equation}\label{e4.03}
 e_{\ominus a}(t,0)\to 0,\; \mbox{as}\; t\to \infty,
\end{equation}
there is an $\alpha > 0$ such that
\begin{gather}\label{e4.04}
    E_1+ \int^t_{0}\!\Big[|a(u)|E_1 + E_2 + E_3\Big]e_{\ominus a}(t,u)
    \, \Delta u \leq \alpha < 1, t \geq 0, \\
\label{e4.05}
    t - g(t) \to \infty, \mbox{as}\; t\to \infty, \\
\label{e4.06}
    Q(t, 0) \to 0, \mbox{as}\; t\to \infty.
\end{gather}


\begin{theorem}\label{thm4.1}
If \eqref{e3.03}--\eqref{e3.04} and \eqref{e4.03}--\eqref{e4.06}
hold, then every solution $x(t, 0, \psi)$ of \eqref{eq1} with small
continuous initial function $\psi (t)$, is bounded and goes to zero
as $t \to \infty$. Moreover, the zero solution is stable at
$t_0=0$.
\end{theorem}

\begin{proof} Define the mapping $P:S \to S$ by
$$
\big(P\varphi\big)(t)= \psi (t)\quad \mbox{if } t \leq 0
$$
and
\begin{align*}
    \big(P\varphi\big)(t)
&=Q(t, \varphi(t-g(t)))+ \Big(\psi(0)
    - Q(0,\psi(-g(0)))\Big)e_{\ominus a}(t,0)\\
& \quad + \int^t_{0}\Big[-a(u)Q^\sigma(u,\varphi(u-g(u)))\\
& \quad + G(u, \varphi(u),\varphi(u-g(u)))\Big] e_{\ominus
    a}(t,u) \, \Delta u, \quad \mbox{if  } t \geq 0.
\end{align*}
It is clear that for $\varphi \in S$, $P\varphi$ is continuous.
Let $\varphi \in S$ with $\|\varphi\|\leq K$, for some positive
constant $K$. Let $\psi(t)$ be a small given continuous initial
function with $\|\psi\|<\delta, \delta >0$. Then,
\begin{equation}\label{e4.7}
\begin{aligned}
    \|P\varphi\|
&\leq  E_1 K+ \Big|\Big(\psi(0) - Q\big(0,
    \psi(-g(0))\big)\Big)\Big|e_{\ominus a}(t,0)\\
& \quad + \int^t_{0}\!\Big[|a(u)|E_1 + E_2 + E_3\Big]e_{\ominus a}(t,u)
    \, \Delta u\;K\\
&\leq \Big(1+ E_1\Big)\delta +E_1 K +
    \int^t_{0}\!\Big[|a(u)|E_1 + E_2 + E_3\Big]e_{\ominus a}(t,u) \,
    \Delta u\;K\\
&\leq \Big(1 + E_1\Big)\delta +  \alpha K,
\end{aligned}
\end{equation}
which implies that, $\|P\varphi\| \leq K$, for the right $\delta$.
Thus, \eqref{e4.7} implies that $(P\varphi)(t)$ is bounded. Next we
show that $\big(P\varphi\big)(t) \to 0$ as $t \to
\infty$. The second term on the right side of $\big (P\varphi
\big)(t)$ tends to zero, by condition \eqref{e4.03}. Also, the first
term on the right side tends to zero, because of \eqref{e4.05},
\eqref{e4.06}  and the fact that $\varphi \in S$. It is left to show
that the integral term goes to zero as $t\to \infty$.

Let $\varepsilon >0$ be given and $\varphi \in S$ with
$\|\varphi\|\leq K$, $K > 0$. Then, there exists a $t_{1} >0$ so
that for $t> t_{1}$, $|\varphi(t-g(t))| < \varepsilon$. Due to
condition \eqref{e4.03}, there exists a $t_{2} > t_{1}$ such that
for $t>t_{2}$ implies that $e_{\ominus a}(t,t_1) <
\frac{\varepsilon}{\alpha K}$.

Thus for $t>t_{2}$, we have
\begin{align*}
 & \Big|\int^t_{0} \!\Big [ -a(u)Q^\sigma\big(u,\varphi(u-g(u))\big)
    + G\big(u,\varphi(u),\varphi(u-g(u))\big)\Big] e_{\ominus a}(t,u) \,
    \Delta u\Big|\\
 & \leq K \int^{t_{1}}_{0} \! \Big[|a(u)|E_1 + E_2 +
    E_3\Big]e_{\ominus a}(t,u) \, \Delta u\\
 & \quad+ \varepsilon \int^t_{t_{1}} \! \Big[|a(u)|E_1 + E_2 +
    E_3\Big]e_{\ominus a}(t,u) \, \Delta u\\
 &\leq Ke_{\ominus a}(t,t_1) \int^{t_{1}}_{0} \!
    \Big[|a(u)|E_1 + E_2 + E_3\Big]e_{\ominus a}(t_1,u) \, \Delta u
    + \alpha \varepsilon\\
 &\leq \alpha Ke_{\ominus a}(t,t_1) + \alpha \varepsilon\\
 &\leq \varepsilon +\alpha \varepsilon.
\end{align*}
Hence, $\big(P\varphi\big)(t) \to 0$ as
$t \to \infty$. It remains to show that $P$ is a
contraction under the supremum norm. Let $\zeta, \eta \in S$. Then
\begin{align*}
 \Big|(P\zeta)(t)-(P\eta)(t)\Big|
&\leq \Big\{E_1 +
\int^t_{0}\Big[|a(u)|E_1 + E_2 + E_3\Big]e_{\ominus a}(t,u) \Delta
u\Big\}\|\zeta - \eta\|\\
&\leq \alpha \|\zeta - \eta\|.
\end{align*}
Thus, by the contraction mapping principle, $P$ has a unique fixed
point in $S$ which solves \eqref{eq1}, is bounded and tends to zero
as $t$ tends to infinity. The stability of the zero solution at $t_0
=0$ follows from the above work by simply replacing $K$ by
$\varepsilon$. This completes the proof.
\end{proof}

\begin{example} \label{exa4.2} \rm
Let
\[
    \mathbb{T} = -2{^\mathbb{Z}} \cup 2^{\mathbb{Z}} \cup \{0\} =
    \{ \dots, -2, -1, -\frac{1}{2}, \dots , 0, \dots,
    \frac{1}{2}, 1, 2, \dots \}.
\]
and let $\psi$ be a continuous initial function, $\psi :
-2^{\mathbb{Z}} \cup \{0\} \to \mathbb{R}$ with $\|\psi\| \leq
\delta$ for small $\delta > 0$. For small $\varepsilon_1$ and
$\varepsilon_2$, we consider the nonlinear neutral dynamic
equation
\begin{equation}\label{e4.08}
    x^\Delta (t) = -2x^\sigma(t) + \varepsilon_1 \big[x^2 (t -
    (t/2) ) \big]^\Delta + \varepsilon_2\big(x^2(t) + x^2(t
    - (t/2))\big).
\end{equation}
The function $g$ is given by $g(t) = \frac{t}{2}$ and satisfies
$g: \mathbb{T} \to \mathbb{R}$ and $id  - g: \mathbb{T} \to
\mathbb{T}$ is strictly increasing.

Suppose
\begin{equation}\label{e4.09}
    0 < 4(2\varepsilon_1 + \varepsilon_2)\delta(1 + \varepsilon_1
    \delta) <1.
\end{equation}
Let $S$ be defined by
\begin{equation*}
    S=\Big\{\varphi :\mathbb{T}\to \mathbb{R}|\;
    \varphi(t)=\psi(t) \mbox{ if } t\leq 0, \varphi(t)\to 0
    \mbox{ as } t\to \infty, \varphi \in C \mbox{ and }
    \|\varphi\| \leq K\Big\},
\end{equation*}
for some positive constant $K$ satisfying the inequality
\begin{equation}\label{e4.10}
\frac{1 - \sqrt{1-4(2\varepsilon_1 + \varepsilon_2)\delta(1 +
\varepsilon_1\;\delta)}}{2(2\varepsilon_1 + \varepsilon_2)} < K <
\frac{1}{2(2\varepsilon_1 + \varepsilon_2)}.
\end{equation}
Then every solution $x(t, 0, \psi)$ of \eqref{e4.08} is bounded
and goes to 0 as $t \to \infty$.
\end{example}

\begin{proof} It is clear from \eqref{e4.09} that inequality \eqref{e4.10}
is well defined. Define
\[
    \big(P\varphi\big)(t)= \psi(t), \, t \leq 0,
\]
and for $t \geq 0$,
\begin{align*}
    \big(P\varphi\big)(t)&=\varepsilon_1 \varphi^2(t/2)+
    \Big(\psi(0)- \varepsilon_1 \psi^2(0)\Big)e_{\ominus 2}(t,
    0)\\
    &\quad + \int^t_{0} \! \Big(-2\varepsilon_1
    \varphi^2\big(\sigma(u)/2\big) + \varepsilon_2 \big(
    \varphi^2(u) + \varphi^2(u/2) \big)\Big) e_{\ominus 2}(t, u)
    \, \Delta u\,.
\end{align*}
Let $\varphi \in S$. Then $\|\varphi\| \leq K$ and for $t \geq 0$,
\begin{align*}
    |\big(P\varphi\big)(t)|
&\leq \varepsilon_1|\varphi(t/2)|^2 +
    \Big( |\psi(0)| + \varepsilon_1|\psi(t/2)|^2\Big) e_{\ominus
    2}(t, 0)\\
&\quad + \int_0^t \! \Big (2 \varepsilon_1 \big | \varphi \big
    ( \sigma(u)/2 \big ) \big |^2 + \varepsilon_2 \big( |
    \varphi(u) |^2 + | \varphi(u/2) |^2 \big)\Big) e_{\ominus
    2}(t, u) \, \Delta u.
\end{align*}
Hence
\begin{equation}\label{e4.11}
\begin{aligned}
    \|P\varphi\| &\leq \varepsilon_1 K^2 + (1
    +\varepsilon_1 \delta) \delta +
    2(\varepsilon_1 + \varepsilon_2)K^2 \int^t_0 \!e_{\ominus
    2}(t, u) \, \Delta u\\
    &\leq (2\varepsilon_1 + \varepsilon_2) K^2 + \delta(1 +
    \varepsilon_1 \delta).
\end{aligned}
\end{equation}
 From \eqref{e4.11}, we see that if,
\begin{equation}\label{e4.12}
    (2\varepsilon_1 + \varepsilon_2)K^2 + \delta(1 +
    \varepsilon_1\;\delta) \leq K,
\end{equation}
then $P: S \to S$. But inequality \eqref{e4.12} is satisfied for
$K$ satisfying \eqref{e4.10}, by noting that
\[
    \frac{1}{2(2\varepsilon_1 + \varepsilon_2)}< \frac{1 +
    \sqrt{1-4(2\varepsilon_1 + \varepsilon_2)\delta(1 +
    \varepsilon_1\;\delta)}}{2(2\varepsilon_1 + \varepsilon_2)}.
\]
Thus, if $\varphi \in S$, then $\|P \varphi\| \leq K$. It is
obvious that conditions \eqref{e4.03}, \eqref{e4.05} and
\eqref{e4.06} are satisfied.

We now show that $P$ defines a contraction mapping on the metric
space $S$. Let $\zeta, \eta \in S$. Then
\begin{align*}
    \Big | (P\zeta)(t) - (P\eta)(t) \Big|
&\leq 2\varepsilon_1 K
    \|\zeta - \eta \| + \Big ( 4 \varepsilon_1 K + 4 \varepsilon_2
    K \Big )\int^t_0 \! e_{\ominus 2}(t, u) \, \Delta u \; \|\zeta
    - \eta\|\\
& \leq \Big [ 2\varepsilon_1 K + 2K(\varepsilon_1 +
    \varepsilon_2 ) \Big ] \|\zeta - \eta\|\\
& =  2(2\varepsilon_1 + \varepsilon_2 ) K \|\zeta - \eta\|.
\end{align*}
By condition \eqref{e4.10}, we have
\[
    \big \|P\zeta - P\eta \big \| \leq \alpha \|\zeta - \eta \|,
\]
for some $\alpha\in (0,1)$. By Theorem \ref{thm4.1}, every solution
$x(\cdot,0,\psi)$ of \eqref{e4.08} with small continuous initial
function $\psi : -2^{\mathbb{Z}} \cup \{0\} \to \mathbb{R}$ is in
$S$, is bounded and goes to zero as $t \to \infty$.
\end{proof}

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\end{document}